Two mutually perpendicular insulated conducti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field $B$ due to an infinitely long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the

Vedclass · Accepted Answer

$\frac{\mu_0 I}{12 \pi}$

Vedclass · Answer

(C) The magnetic field $B$ due to an infinitely long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the $x$-axis,the perpendicular distance to point $P(2, 3)$ is $d_x = 3 \ m$. Using the right-hand thumb rule,the direction of the magnetic field at $P$ is outward (denoted by $\odot$).$B_x = \frac{\mu_0 I}{2 \pi (3)} = \frac{\mu_0 I}{6 \pi} \odot$For the wire along the $y$-axis,the perpendicular distance to point $P(2, 3)$ is $d_y = 2 \ m$. Using the right-hand thumb rule,the direction of the magnetic field at $P$ is inward (denoted by $\otimes$).$B_y = \frac{\mu_0 I}{2 \pi (2)} = \frac{\mu_0 I}{4 \pi} \otimes$The net magnetic field $B_{	ext{net}}$ is the difference between the two fields because they are in opposite directions:$B_{	ext{net}} = B_y - B_x = \frac{\mu_0 I}{4 \pi} - \frac{\mu_0 I}{6 \pi}$$B_{	ext{net}} = \frac{3 \mu_0 I - 2 \mu_0 I}{12 \pi} = \frac{\mu_0 I}{12 \pi}$

Two mutually perpendicular insulated conducting wires carrying equal currents $I$ intersect at the origin. The resultant magnetic induction at point $P(2 \ m, 3 \ m)$ will be:

Similar Questions

Identical currents flow in two perpendicular wires,as shown in the figure. The wires are very close but do not touch. The magnetic field can be zero:

Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?

In the diagram,$I_{1}$ and $I_{2}$ are the strengths of the currents in the loop and straight conductor,respectively. Given $OA = AB = R$. The net magnetic field at the centre $O$ is zero. Then the ratio of the currents in the loop and the straight conductor is:

$A$ cell is connected between two points of a uniformly thick circular conductor. The magnetic field at the centre of the loop will be

$A$ direct current is passing through a wire. It is bent to form a coil of one turn. Now,it is further bent to form a coil of two turns but at a smaller radius. Find the ratio of the magnetic induction at the centre of this two-turn coil to that at the centre of the one-turn coil.

Vedclass Test Series

Exam Paper Generator

Online Exam Module